Combination of Resistors - Series and Parallel Combination

A wire of  $20\Omega$  resistance is gradually stretched to double its original length. It is then cut into two equal parts. These parts are then connected in parallel across a 4.0 volt battery. The current drawn from the battery is:

A (i) series (ii) parallel combination of two given resistors is connected, one – by – one, across a cell. In which will the terminal potential difference, across the cell, have a higher value?

(i) Calculate the equivalent resistance of the given electrical network between points $A$ and $B$.

(ii) Also calculate the current through $CD$, if a $10 \mathrm{V} DC$ source is connected between $A$ and $B$, and the value of $R$ is assumed as  $2\Omega$.

Find the equivalent resistance between point $A$ and $B$

What will be the most suitable combination of three resistors $A=2 \mathrm{\Omega }, B=4 \mathrm{\Omega }, C=6 \mathrm{\Omega }$ so that $\left(\frac{22}{3}\right)\mathrm{\Omega }$ is equivalent resistance of combination?

For the circuit given below, what is the effective resistance between points $a$ and $b$

If $400\Omega$ of resistance is made by adding four $100\Omega$ resistance of tolerance $5\%$ then the tolerance of the combination in the percentage is.

The figure shows a network in which the cell is ideal and it has an emf $E$. The potential difference across the resistance $2R$ is

In the given circuit all resistance are of value of $R$ ohm each. The equivalent resistance between $A$ and $B$ is:

The figure shows three circuits $\mathrm{I},\mathrm{ }\mathrm{II}$ and $\mathrm{III}$ which are connected to a $3 \mathrm{V}$ battery. If the powers dissipated by the configurations $\mathrm{I},\mathrm{ }\mathrm{II}$ and $\mathrm{III}$ are $P_1, P_2$ and $P_3$ respectively, then -

In the shown wireframe, each side of a square (the smallest square) has a resistance $7 \mathrm{\Omega }$. What is the equivalent resistance value (in ohm) of the circuit between the points $\mathrm{A}$ and $\mathrm{B}$?

In the circuit in figure we have ${\epsilon }_1={\epsilon }_2=100 V$, $R_1=20\Omega ,R_2=10\Omega ,R_3=40\Omega$ and $R_4=30\Omega$ Find the reading of the ammeter (in $A$). Disregard the resistance of the battery and the ammeter.

The equivalent resistance between $A$ and $B$ for the given circuit is $20x$ ohm where, $R=100\sqrt{3} \mathrm{\Omega }$. Value of $x$ is,

In the figure shown, segments $AB, BC, CD$ and $DA$, as well as the diagonal segments $AC$ and $BD$ have a resistance $2\Omega$ each. The mid points of the diagonal wires are soldered with each other. If the effective resistance between the mid points of sides $AD$ and $CD$ is $\frac{x}{4}\Omega$, find the value of $x$.

Resistance of each part is $\text{'}R\text{'}$ and resistance of circumference is negligible as shown in figure. The equivalent resistance across $AB$ is $\frac{3R}{\alpha }.$ Fill the value of $\alpha .$

A wire of uniform cross-section and resistance $80 \mathrm{\Omega }$ is bent in the form of a square $ABCD.$ Point $A$ is connected to a point $P$ on $DC$ by a wire $AP$ of resistance $20 \mathrm{\Omega }.$ When potential difference is applied between $A$ and $C,$ the points $B$ and $P$ are seen to be at the same potential. The resistance of part DP is $x \mathrm{\Omega }.$Find the value of $x\mathit{.}$ (Approximate the answer to the nearest integer)

The equivalent resistance between $A$ and $B$ is $(\alpha /2)\Omega$. The value of $\alpha$ is

Four infinite ladder alike networks are combined as shown in figure .Each resistance is $R\Omega$. The equivalent resistance between $A$ and $B$ is $R_{\mathrm{\text{AB}}}$ and between $A$ and $C$ is $R_{\mathrm{\text{AC}}}.$ Let ratio $\frac{R_{\mathrm{\text{AB}}}}{R_{\mathrm{\text{AC}}}}\mathrm{\text{is}}\frac{x}{4}$ then find the value of $x$.